【leetcode】24.两两交换链表中的节点

25题是每k个reverse，这一题是每2个reverse，这里就可以直接拿25题的代码用了

使用reverse方法

var swapPairs = function (head) {
    if (!head) return head
    let left = head, right = head
    for (let i = 0; i < 2; ++i) {
        // 不足两个就是到尾部了，直接衔接即可
        if (!right) return head
        right = right.next
    }
    // 反转后得到的新链表
    let list = reverse(left, right)
    // 将新链表与后面的衔接
    left.next = swapPairs(right)
    return list
};
// 区间内进行reverse
var reverse = function (left, right) {
    let pre = null, current = left
    while (current !== right) {
        let currentNext = current.next
        current.next = pre
        pre = current
        current = currentNext
    }
    return pre
}

递归

递归的思路就比较简单一些了

var swapPairs = function (head) {
    if (!head || !head.next) return head
    let nextHead = head.next
    head.next = swapPairs(nextHead.next)
    nextHead.next = head
    return nextHead
};

迭代

这里需要设置头节点dummy，方便后续返回链表头

var swapPairs = function (head) {
    if (!head) return head
    let dummy = new ListNode(-1)
    dummy.next = head
    let current = dummy
    while (current.next && current.next.next) {
        let first = current.next
        let second = current.next.next
        current.next = second
        first.next = second.next
        second.next = first
        // 挪动current节点
        current = first
    }
    return dummy.next
};